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I am designing a microprocessor driven wristwatch with a VFD tubes 2 digits display. These are the three major requirements:

  • VFDs operate at 15V
  • uC operates at 3V
  • I have to use as less space as possible

My plan is to use a 3V battery and a dc to dc step-up converter to get the 15V when I need them (that is, when I decide I want to display the time on the two VFDs). Now I have to select a suitable integrated circuit, which fits my very special needs, and this is something I've never done. So, these are the specs I have for the booster:

  • Vin 3V
  • Vout 15V
  • Iout 300mA

Based on this parameters, I've identified the following component: LT3495

Do you think it would be suitable for what I need? Does a designer normally search for a new component this way, or normally are more specifications, more details needed?

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    \$\begingroup\$ 15V 300mA, this makes 4.5W. If you really need that, I'm afraid the 3V button cell won't last long. \$\endgroup\$ – dim Dec 23 '16 at 16:16
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    \$\begingroup\$ Ditto what @dim says so please provide link to the battery and the display because 300 mA seems mighty high. \$\endgroup\$ – Andy aka Dec 23 '16 at 16:17
  • \$\begingroup\$ My plan was to use those 3v lithium batteries for cameras... The VFD won't be on all the time. Most of the time they'll be switched off. Like this one: dubuque-forsale.com/This-Month-Ads/Kids-clothing/Boys/… \$\endgroup\$ – Enrico Dec 23 '16 at 16:19
  • \$\begingroup\$ I am designing the display. I'll use two of these: blog-imgs-24-origin.fc2.com/a/r/m/arms22/P1010420_a.jpg \$\endgroup\$ – Enrico Dec 23 '16 at 16:20
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    \$\begingroup\$ Why do you think you need 300mA out? \$\endgroup\$ – user_1818839 Dec 23 '16 at 16:21
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300 mA out at 15 volts is a power of 4.5 watts and, with the inefficiency of the converter you will need to supply about 5 watts from the 3 volt battery (or 1.67 amps). However, the output resistance of a Duracell CR17345 is about 0.25 ohms and taking this amount of current instantly drops the battery terminal voltage from 3 volts to about 2.6 volts. You are now on the slippery slope of it never working - the voltage drops and your DC-DC converter demands even more current to get 5 watts it needs to sustain 15 volts on the output at 300 mA.

enter image description here

See how the battery terminal voltage has dropped to about 2.6 volts with a load current of just 1 amp after only 6 minutes - given the output resistance is 0.25 ohms, a 1 amp load will immediately cause a voltage drop to 2.75 volts.

The LT3495 requires an input voltage of at least 2.5 volts so after a few tens or maybe hundreds of milli seconds it will shut down then burst back to life (as the load is removed due to the shut-down) then continue repeating this and flashing the display on and off cyclically.

You need to find a display technology that isn't so hungry or, have a better battery.

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  • \$\begingroup\$ Thank you for the very interesting explanation. You and @Brian have given me the answer i needed. I'll review the design of the display, and try to make it less power demanding. I really don't want a bigger battery. \$\endgroup\$ – Enrico Dec 23 '16 at 16:54
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Here are some helpful considerations I've dealt with when choosing a boost converter. and specifically your project. This is more directed at the whole boost converter system and not just the controller chip managing the boost.

  1. efficiency, quiescent drain. You can't get away from this top priority in any device using a battery. Even given a very high efficiency converter, that efficiency must drop to 0% at no load. So more important than the 15 volts you need output, is how well the booster is matched to the task. It may be obvious, but in my experience any boost circuit is typically most efficient when used somewhere around 60% of its capacity or more. Especially if there are times when the boost circuit will be on and its output unused (such as after timeout on a backlight), you'll want to pay great attention to the input current at that no load condition, and be sure you're OK with this addition to the battery drain. Specifically, that LT3495 does look like a reasonable choice, though as already pointed out your current demand will likely make your choice of such a small power cell impractical, unless the load will only be needed very briefly. You can't hope for more mA-hours than the battery provides.

  2. Minimum operational voltage. You are treading a fine enough line building a boost converter to operate at 3V, and you need to carefully consider both the operational drop out point, and how the likely drop in efficiency might hasten the battery drain as the voltage decreases. This is not an impossible task. Considering how many devices operate for long periods from just a 1.5V standard cell. Such a cell might be a better choice mA-hour wise, but for that you'll need a different building block than the LT3495. Perhaps consider upgrading the design to a small rechargeable Li-Ion cell if space permits.

  3. EMI and noise on the output. I recently had a very challenging time with a somewhat higher current boost converter used to drive an audio amplifier, in a situation where it was in close proximity to a magnetic guitar pickup. Needless to say I got a major education about a seldom discussed issue with boost converters: They are VERY noisy! Even if you're not dealing with something as sensitive as audio, you may need to take into account the radiated EMI noise, and the somewhat surprisingly tall current and voltage spikes a boost converter may add to the output. I've also found that the greater the boost ratio, the messier these spikes can become. Depending on the sensitivity of your load, you may need to add additional filtering on the output, and sometimes even shielding around the converter itself. Going back a step, you can minimize these problems and likely increase efficiency by choosing components ideal to the task. Examples include a VERY LOW ERSR capacitor for the output, and a torrid type inductor.

Best of luck!

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